If the line `a x+b y=2` is a normal to the circle `x^2+y^2-4x-4y=0` and a tangent to the circle `x^2+y^2=1` , then

Options

A. `a=(1)/(2),b=(1)/(2)`
B. `a=(1+sqrt(7))/(2),b=(1-sqrt(7))/(2)`
C. `a=(1)/(4),b=(3)/(4)`
D. `a=1,b=sqrt(3)`

Answer

2

Explanation

The center of `x^(2)+y^(2)-4x-4y=0` is (2,2)
Given normal is `ax+by=2`. Therefore,
`2a+2b=2` or `a+b=1`
`ax+by=2` touches `x^(2)+y^(2)=1`. So,
`1=|(-2)/(sqrt(a^(2)+b^(2)))|`
`:. a^(2)+b^(2)=4` or `a^(2)+(1-a^(2))=4`
or `2a^(2)-2a-3=0`
`:. a =(2+- sqrt(4+24))/(4)=(1+- sqrt(7))/(2)`
`:. b= 1-a= 1 -(1+-sqrt(7))/(2)=(1+- sqrt(7))/(2)`